I'm not sure that i follow this. Is it based on any particular test? How do you arrive at that conclusion? I'm sorry for asking, but it seems that i messed up my understanding of sequences v/s series. I can relate this to AST, where the limit of [itex]a_n[/itex] needs to be zero for the series to converge.

So for it to converge the limit has to go to zero ? (Don't think the contrapostive of the non-null test is true though ?)

The contrapositive has to be true. It's logically equivalent to the original statement. The nth-term test says if ##\displaystyle \lim_{n\to\infty}a_n\ne 0## then the infinite series ##\displaystyle \sum_n a_n## will not converge. The contrapositive would be: if the series converges, then the limit of an has to be 0.

What I think you're thinking is if an goes to 0, it doesn't necessarily mean that the series converges. That is correct.

Say I put [tex] \sum[({1-\frac{1}{r}})^r]^2 [/tex]

Than took the limit of the inner part would I get [tex] \frac{1}{e^2} [/tex] It still would not be equal to zero...?

I'm not sure that i follow this. Is it based on any particular test? How do you arrive at that conclusion? I'm sorry for asking, but it seems that i messed up my understanding of sequences v/s series. I can relate this to AST, where the limit of [itex]a_n[/itex] needs to be zero for the series to converge.

You've forgotten about the nth-term test. It's probably the very first test mentioned when you began to study series.

You've forgotten about the nth-term test. It's probably the very first test mentioned when you began to study series.

I know about the nth-term test but its definition says that as long as the limit of the sequence does not equal zero, the series has to diverge. This theorem says nothing about convergence. If the limit is equal to zero, then according to the nth-term test, the series could either converge or diverge.

The contrapositive has to be true. It's logically equivalent to the original statement. The nth-term test says if ##\displaystyle \lim_{n\to\infty}a_n\ne 0## then the infinite series ##\displaystyle \sum_n a_n## will not converge. The contrapositive would be: if the series converges, then the limit of an has to be 0.

I'm getting confused as well, an obvious example would be:

[tex] a_n = \frac{1}{n} -> 0 [/tex] as n->infinity

But

[tex] \sum \frac{1}{n} [/tex]

diverges ?

So how is the contrapositive true.

If your saying that we already know that the series is converging than that statement is true, that maybe different but in this case we don't know if it converges or diverges.